Physicists tell us that altitude h in feet of a projectilet seconds after firing is h=-1672 +8/+he,where y, is the initial velocity in feet per second and h, is the altitude in feet from which it is fired. If arocket is launched from a hilltop 2400 feet above the desert with an initial upward velocity of 400 feet persecond, then when will it land on the desert?

Question

contestada

Respuesta :

DezireW8312 DezireW8312 · Answer 1 · 2022-11-03T04:43:10+01:00

The projectile will land on the desert when h = 0, in other words

[tex]0=-16t^2+v_0t+h_0[/tex]

Now, we are told that h_o = 2400ft and v_o = 400ft/s; and putting those in the formula above gives

[tex]-16t^2+400_{}t+2400=0[/tex]

Now this is a quadratic equation and the solution is given by

[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]

In our case, a = -16, b = 400, and c = 2400; therefore, the quadratic formula above gives

[tex]t=\frac{-400\pm\sqrt[]{400^2-4(-16)(2400)}}{2(-16)}[/tex][tex]t=\frac{-400\pm560}{-32}[/tex]

Hence, the two solutions are

[tex]\begin{gathered} t=30 \\ t=-5 \end{gathered}[/tex]

Since negative values of t have no physical significance, the valid answer is t = 30s.

Hence, the third choice in the column is the right answer.